递归列表展平

Question

我自己可以写这个，但是我想要完成它的具体方法就是把我扔掉。我正在尝试编写一个类似于.NET 3.5中引入的其他通用扩展方法，它将采用IEnumerables的嵌套IEnumerable（依此类推）并将其展平为一个IEnumerable。有人有什么想法吗？

具体来说，我遇到了扩展方法本身的语法问题，因此我可以使用展平算法。

Answer 1

这是一个可能有帮助的扩展。它将遍历对象层次结构中的所有节点，并选择符合条件的节点。它假定层次结构中的每个对象都有一个包含其子对象的集合属性。

这是扩展名：

/// Traverses an object hierarchy and return a flattened list of elements
/// based on a predicate.
/// 
/// TSource: The type of object in your collection.</typeparam>
/// source: The collection of your topmost TSource objects.</param>
/// selectorFunction: A predicate for choosing the objects you want.
/// getChildrenFunction: A function that fetches the child collection from an object.
/// returns: A flattened list of objects which meet the criteria in selectorFunction.
public static IEnumerable<TSource> Map<TSource>(
  this IEnumerable<TSource> source,
  Func<TSource, bool> selectorFunction,
  Func<TSource, IEnumerable<TSource>> getChildrenFunction)
{
  // Add what we have to the stack
  var flattenedList = source.Where(selectorFunction);

  // Go through the input enumerable looking for children,
  // and add those if we have them
  foreach (TSource element in source)
  {
    flattenedList = flattenedList.Concat(
      getChildrenFunction(element).Map(selectorFunction,
                                       getChildrenFunction)
    );
  }
  return flattenedList;
}

示例（单元测试）：

首先，我们需要一个对象和一个嵌套的对象层次结构。

一个简单的节点类

class Node
{
  public int NodeId { get; set; }
  public int LevelId { get; set; }
  public IEnumerable<Node> Children { get; set; }

  public override string ToString()
  {
    return String.Format("Node {0}, Level {1}", this.NodeId, this.LevelId);
  }
}

一种获得3级深度节点层次结构的方法

private IEnumerable<Node> GetNodes()
{
  // Create a 3-level deep hierarchy of nodes
  Node[] nodes = new Node[]
    {
      new Node 
      { 
        NodeId = 1, 
        LevelId = 1, 
        Children = new Node[]
        {
          new Node { NodeId = 2, LevelId = 2, Children = new Node[] {} },
          new Node
          {
            NodeId = 3,
            LevelId = 2,
            Children = new Node[]
            {
              new Node { NodeId = 4, LevelId = 3, Children = new Node[] {} },
              new Node { NodeId = 5, LevelId = 3, Children = new Node[] {} }
            }
          }
        }
      },
      new Node { NodeId = 6, LevelId = 1, Children = new Node[] {} }
    };
  return nodes;
}

第一次测试：展平层次结构，不过滤

[Test]
public void Flatten_Nested_Heirachy()
{
  IEnumerable<Node> nodes = GetNodes();
  var flattenedNodes = nodes.Map(
    p => true, 
    (Node n) => { return n.Children; }
  );
  foreach (Node flatNode in flattenedNodes)
  {
    Console.WriteLine(flatNode.ToString());
  }

  // Make sure we only end up with 6 nodes
  Assert.AreEqual(6, flattenedNodes.Count());
}

这将显示：

Node 1, Level 1
Node 6, Level 1
Node 2, Level 2
Node 3, Level 2
Node 4, Level 3
Node 5, Level 3

第二次测试：获取具有偶数NodeId的节点列表

[Test]
public void Only_Return_Nodes_With_Even_Numbered_Node_IDs()
{
  IEnumerable<Node> nodes = GetNodes();
  var flattenedNodes = nodes.Map(
    p => (p.NodeId % 2) == 0, 
    (Node n) => { return n.Children; }
  );
  foreach (Node flatNode in flattenedNodes)
  {
    Console.WriteLine(flatNode.ToString());
  }
  // Make sure we only end up with 3 nodes
  Assert.AreEqual(3, flattenedNodes.Count());
}

这将显示：

Node 6, Level 1
Node 2, Level 2
Node 4, Level 3

Answer 2

嗯......我不确定完全你想要什么，但这里是“一级”选项：

public static IEnumerable<TElement> Flatten<TElement,TSequence> (this IEnumerable<TSequence> sequences)
    where TSequence : IEnumerable<TElement> 
{
    foreach (TSequence sequence in sequences)
    {
        foreach(TElement element in sequence)
        {
            yield return element;
        }
    }
}

如果那不是你想要的，你能提供你想要的签名吗？如果您不需要通用表单，并且您只想做LINQ to XML构造函数所做的事情，那就相当简单 - 尽管迭代器块的递归使用效率相对较低。类似的东西：

static IEnumerable Flatten(params object[] objects)
{
    // Can't easily get varargs behaviour with IEnumerable
    return Flatten((IEnumerable) objects);
}

static IEnumerable Flatten(IEnumerable enumerable)
{
    foreach (object element in enumerable)
    {
        IEnumerable candidate = element as IEnumerable;
        if (candidate != null)
        {
            foreach (object nested in candidate)
            {
                yield return nested;
            }
        }
        else
        {
            yield return element;
        }
    }
}

请注意，这会将字符串视为字符序列，但是 - 您可能希望特殊字符串是单个元素而不是展平它们，具体取决于您的用例。

这有帮助吗？

Answer 3

我认为我会分享一个完整的错误处理示例和单逻辑应用程序。

递归展平非常简单：

LINQ版

public static class IEnumerableExtensions
{
    public static IEnumerable<T> SelectManyRecursive<T>(this IEnumerable<T> source, Func<T, IEnumerable<T>> selector)
    {
        if (source == null) throw new ArgumentNullException("source");
        if (selector == null) throw new ArgumentNullException("selector");

        return !source.Any() ? source :
            source.Concat(
                source
                .SelectMany(i => selector(i).EmptyIfNull())
                .SelectManyRecursive(selector)
            );
    }

    public static IEnumerable<T> EmptyIfNull<T>(this IEnumerable<T> source)
    {
        return source ?? Enumerable.Empty<T>();
    }
}

非LINQ版本

public static class IEnumerableExtensions
{
    public static IEnumerable<T> SelectManyRecursive<T>(this IEnumerable<T> source, Func<T, IEnumerable<T>> selector)
    {
        if (source == null) throw new ArgumentNullException("source");
        if (selector == null) throw new ArgumentNullException("selector");

        foreach (T item in source)
        {
            yield return item;

            var children = selector(item);
            if (children == null)
                continue;

            foreach (T descendant in children.SelectManyRecursive(selector))
            {
                yield return descendant;
            }
        }
    }
}

设计决策

我决定：

• 禁止展开null IEnumerable，可以通过删除异常抛出来改变它：
  • 在第一个版本的source = source.EmptyIfNull();之前添加return
  • 在第二个版本的if (source != null)之前添加foreach
• 允许选择器返回null集合 - 这样我就会从调用者中删除责任，以确保子列表不为空，这可以通过以下方式更改：
  • 删除第一个版本中的.EmptyIfNull() - 请注意，如果选择器返回null，则SelectMany将失败
  • 删除第二个版本中的if (children == null) continue; - 请注意，foreach参数<{1}} IEnumerable将失败
• 允许在调用者端或子选择器中过滤带有.Where子句的子项，而不是传递子过滤器选择器参数：
  • 它不会影响效率，因为在两个版本中都是延迟调用
  • 它会将另一个逻辑与方法混合，我更喜欢将逻辑分开

使用示例

我在LightSwitch中使用此扩展方法来获取屏幕上的所有控件：

public static class ScreenObjectExtensions
{
    public static IEnumerable<IContentItemProxy> FindControls(this IScreenObject screen)
    {
        var model = screen.Details.GetModel();

        return model.GetChildItems()
            .SelectManyRecursive(c => c.GetChildItems())
            .OfType<IContentItemDefinition>()
            .Select(c => screen.FindControl(c.Name));
    }
}

Answer 4

这不是[SelectMany] [1]的用途吗？

enum1.SelectMany(
    a => a.SelectMany(
        b => b.SelectMany(
            c => c.Select(
                d => d.Name
            )
        )
    )
);

Answer 5

这是一个修改后的Jon Skeet's answer，允许超过“一个级别”：

static IEnumerable Flatten(IEnumerable enumerable)
{
    foreach (object element in enumerable)
    {
        IEnumerable candidate = element as IEnumerable;
        if (candidate != null)
        {
            foreach (object nested in Flatten(candidate))
            {
                yield return nested;
            }
        }
        else
        {
            yield return element;
        }
    }
}

免责声明：我不知道C＃。

Python中的相同内容：

#!/usr/bin/env python

def flatten(iterable):
    for item in iterable:
        if hasattr(item, '__iter__'):
            for nested in flatten(item):
                yield nested
        else:
            yield item

if __name__ == '__main__':
    for item in flatten([1,[2, 3, [[4], 5]], 6, [[[7]]], [8]]):
        print(item, end=" ")

打印：

1 2 3 4 5 6 7 8 

Answer 6

功能：

public static class MyExtentions
{
    public static IEnumerable<T> RecursiveSelector<T>(this IEnumerable<T> nodes, Func<T, IEnumerable<T>> selector)
    {
        if(nodes.Any())
            return nodes.Concat(nodes.SelectMany(selector).RecursiveSelector(selector));

        return nodes;
    } 
}

用法：

var ar = new[]
{
    new Node
    {
        Name = "1",
        Chilren = new[]
        {
            new Node
            {
                Name = "11",
                Children = new[]
                {
                    new Node
                    {
                        Name = "111",

                    }
                }
            }
        }
    }
};

var flattened = ar.RecursiveSelector(x => x.Children).ToList();

Answer 7

SelectMany扩展方法已经这样做了。

  

将序列的每个元素投影到   IEnumerable＆lt;（＆lt;（T＆gt;）＆gt;）和   将得到的序列展平成   一个序列。

Answer 8

由于在VB中没有yield，LINQ提供了延迟执行和简洁的语法，你也可以使用。

<Extension()>
Public Function Flatten(Of T)(ByVal objects As Generic.IEnumerable(Of T), ByVal selector As Func(Of T, Generic.IEnumerable(Of T))) As Generic.IEnumerable(Of T)
    Return objects.Union(objects.SelectMany(selector).Flatten(selector))
End Function

Answer 9

我必须从头开始实施我的，因为所有提供的解决方案都会中断，以防有一个循环，即指向其祖先的孩子。如果您有与我相同的要求，请看一下（如果我的解决方案在任何特殊情况下都会中断，也请告诉我）：

使用方法：

var flattenlist = rootItem.Flatten(obj => obj.ChildItems, obj => obj.Id)

代码：

public static class Extensions
    {
        /// <summary>
        /// This would flatten out a recursive data structure ignoring the loops. The end result would be an enumerable which enumerates all the
        /// items in the data structure regardless of the level of nesting.
        /// </summary>
        /// <typeparam name="T">Type of the recursive data structure</typeparam>
        /// <param name="source">Source element</param>
        /// <param name="childrenSelector">a function that returns the children of a given data element of type T</param>
        /// <param name="keySelector">a function that returns a key value for each element</param>
        /// <returns>a faltten list of all the items within recursive data structure of T</returns>
        public static IEnumerable<T> Flatten<T>(this IEnumerable<T> source,
            Func<T, IEnumerable<T>> childrenSelector,
            Func<T, object> keySelector) where T : class
        {
            if (source == null)
                throw new ArgumentNullException("source");
            if (childrenSelector == null)
                throw new ArgumentNullException("childrenSelector");
            if (keySelector == null)
                throw new ArgumentNullException("keySelector");
            var stack = new Stack<T>( source);
            var dictionary = new Dictionary<object, T>();
            while (stack.Any())
            {
                var currentItem = stack.Pop();
                var currentkey = keySelector(currentItem);
                if (dictionary.ContainsKey(currentkey) == false)
                {
                    dictionary.Add(currentkey, currentItem);
                    var children = childrenSelector(currentItem);
                    if (children != null)
                    {
                        foreach (var child in children)
                        {
                            stack.Push(child);
                        }
                    }
                }
                yield return currentItem;
            }
        }

        /// <summary>
        /// This would flatten out a recursive data structure ignoring the loops. The     end result would be an enumerable which enumerates all the
        /// items in the data structure regardless of the level of nesting.
        /// </summary>
        /// <typeparam name="T">Type of the recursive data structure</typeparam>
        /// <param name="source">Source element</param>
        /// <param name="childrenSelector">a function that returns the children of a     given data element of type T</param>
        /// <param name="keySelector">a function that returns a key value for each   element</param>
        /// <returns>a faltten list of all the items within recursive data structure of T</returns>
        public static IEnumerable<T> Flatten<T>(this T source, 
            Func<T, IEnumerable<T>> childrenSelector,
            Func<T, object> keySelector) where T: class
        {
            return Flatten(new [] {source}, childrenSelector, keySelector);
        }
    }

Answer 10

好的，这是另一个版本，由上面的3个答案组合而成。

递归。使用产量。通用。可选的过滤谓词。可选的选择功能。关于我能做到的简洁。

    public static IEnumerable<TNode> Flatten<TNode>(
        this IEnumerable<TNode> nodes, 
        Func<TNode, bool> filterBy = null,
        Func<TNode, IEnumerable<TNode>> selectChildren = null
        )
    {
        if (nodes == null) yield break;
        if (filterBy != null) nodes = nodes.Where(filterBy);

        foreach (var node in nodes)
        {
            yield return node;

            var children = (selectChildren == null)
                ? node as IEnumerable<TNode>
                : selectChildren(node);

            if (children == null) continue;

            foreach (var child in children.Flatten(filterBy, selectChildren))
            {
                yield return child;
            }
        }
    }

用法：

// With filter predicate, with selection function
var flatList = nodes.Flatten(n => n.IsDeleted == false, n => n.Children);

Answer 11

static class EnumerableExtensions
{
    public static IEnumerable<T> Flatten<T>(this IEnumerable<IEnumerable<T>> sequence)
    {
        foreach(var child in sequence)
            foreach(var item in child)
                yield return item;
    }
}

也许是这样的？或者你的意思是它可能无限深？

Answer 12

class PageViewModel { 
    public IEnumerable<PageViewModel> ChildrenPages { get; set; } 
}

Func<IEnumerable<PageViewModel>, IEnumerable<PageViewModel>> concatAll = null;
concatAll = list => list.SelectMany(l => l.ChildrenPages.Any() ? 
    concatAll(l.ChildrenPages).Union(new[] { l }) : new[] { l });

var allPages = concatAll(source).ToArray();

Answer 13

基本上，你需要有一个超出递归函数的主IENumerable，然后是你的递归函数（Psuedo-code）

private void flattenList(IEnumerable<T> list)
{
    foreach (T item in list)
    {
        masterList.Add(item);

        if (item.Count > 0)
        {
            this.flattenList(item);
        }
    }
}

虽然我真的不确定IEnumerable嵌套在IEnumerable中你的意思...那是什么意思？嵌套多少级？什么是最终类型？显然我的代码不正确，但我希望它让你思考。